Phys 954, Solar Wind and Cosmic Rays
E. Möbius
Physics 954
Homework #6
1)
Solutions
Diffuse Energetic Ions at the Earth's bow shock
j = 2.104 s-1 cm-2 sr-1 keV-1
from SW to 20 keV
j  5.4104 e E / Eo u
for E > Eo = 20 keV
msw = 4 cm-3; VSW = 480 km/s;
B = 20 nT
a)
We assume an isotropic distribution in the solar wind frame and compute the density and
energy density of the diffuse ions in the solar wind frame. The density is unaffected by the
frame. We neglect the effects of the transformation on the energy density.

NDi ff   f (v)4 v 2 dv
0
With
j

dJ 
cts

2
dEd 
s cm sr keV 

cm 2 
 cts 
 0.9610 v
f (v)  6
2 

cm / sec3 

sec 
15
we get:
2

4
NDi ff 
j dv
0.961015 0
vo 
NDiff 
2Eo
cm
 4.4107  20
m
sec
4
0.96  1015

 v 
20 4.4 10  j d   [cm 3 ]
vo 
0
7
Use
N Diff [cm
3
] 5.7610
7
  v 
 Eo[keV]  j d  
vo 
0
  v 
 2.510 6  j d  
vo 
0
Phys 954, Solar Wind and Cosmic Rays
E  Eo 
We can express the energy:

 Diff keV cm
3

E. Möbius
5.7810
7
v2
vo 2
and get the energy density

 v 2  v 
Eo [keV] Eo [keV]  j    d  
0 vo  vo 
We have to solve the integrals:

A)


 
v 
d v  2 10 4  d  v 
 5.4 10 4  e
 j 
0 vo  vo
0 vo 
1
1
2
v
v o2
 v 
d 
vo 
for the density and
1  2  
 v
 v  v 2  v 
v
v
4
4
v 2
B)
0 j vo vo  d vo  210 0 vo  dvo  5.410 1 e o
for the energy density?

A)
2
v2   v 

v 
d 
 

 o  vo 
The first integral yields
1  v 
4
4
2 10  d    2 10
0 vo 
the second integral can be broken into
2
2
v

v

  2  v 
  2  v  1
4
v
4
v
5.410  e o d   5.410  e o d    dto
vo 
vo  0 
0
1


4    2 
4
 5.410    
 0.810


e 
 2
The second term is determined by numerical integration or can be estimated in comparison with
the first integral after plotting the functions (experimental physicists thumb rule method!)
The integrals are a combination of an integral over a constant (up to Eo) and over an exponential
for the rest of the energy range.
For the density we get:
NDiff [cm-3] = 2.5.10-6*[2.104 + 5.4.104(√π/2 - 2/e)] ≈ 0.07
i.e. the number density of the diffuse ions is about 1.4% of the SW density.
For the energy density we have to multiply the integrand by (v/vo)2 and get:
Diff [keV/cm-3] = 5.8.10-7.√20.20*[6.6.103 + 5.4.104(√π/4 - 0.19)] ≈ 1.04
Phys 954, Solar Wind and Cosmic Rays
E. Möbius
b) The solar wind ram kinetic energy is
1
2
3
 kin  mp v SW2 nSW 1 keV ( 480
440 )  4cm
2
 5keV / cm 3
The magnetic energy density is
9 2
B 2 20  10 
J
Emag 

 1.610 11 3
7
2 o 2  4  10
m
 1.610 11 106 /1.6 10 19 / 103
keV
keV
 1 3
s
cm
cm
The energy density of this population is ≈ 20% of the ram kinetic energy density of the solar
wind and approximately that of the IMF energy density. The solar bulk energy can still supply
enough energy to drive the energetization of the upstream particles, but the energy density of the
diffuse ions is a significant fraction of the solar wind energy density.
However, the magnetic field will definitely be affected by the energetic particles. The magnetic
field is tied to the SW plasma and thus strengthened against the push of the particles, but it will
be altered. Therefore, a test particle picture is not appropriate in this situation. Indeed the ions
create strong waves in the IMF, and a description that combines waves with the particles in a
self-consistent manner is needed.
c) We found out that pitch-angle scattering is the most effective transport process in
interplanetary space, i.e. it has the smallest mean free path length for scattering. Complete pitchangle scattering of a particle population leads to an isotropic distribution in the plasma frame, in
our case in the frame of the solar wind. If we are talking about a diffusion process in front of the
bow shock we assume near isotropy of the particle population. This is indeed justified, because
the mean free path for pitch-angle scattering is of the order of a few Earth radii near the bow
shock.
Of course, there will also be complete pitch angle scattering to isotropy in the downstream
region, because the fluctuations remain in the plasma. They are even amplified during the shock
passage. However, it will take a short distance from the shock until the energetic distribution is
completely isotropized. This exact distance is still under investigation. Cluster may help here.
Assuming isotropy in both upstream and downstream flow frame leads to a compression of the
distribution, which turns into acceleration. This diffusive acceleration is known as first order
Fermi acceleration. The main processes are energy gain reflection with moving fluctuations and
subsequent by re-distribution of this increased energy to isotropy. So the next reflection can
again contribute energy to the acceleration process.
Phys 954, Solar Wind and Cosmic Rays
2)
E. Möbius
Compton-Getting-Effect
a)
The anisotropies are defined as:
A = (ju – jd)/(ju + jd)
where u and d mean the viewing of the fluxes in the upstream and downstream direction,
respectively. For VsW/v << 1 we may use the approximation;
A  2 VSW/V
where  is the spectral index of the flux spectrum (when represented as a power law). In the
following table the computations are given in the approximation and the full representation. As
can be seen the approximation is only good for E > 100 keV/Nuc. For a nominal solar wind
speed of 440 km/sec (ESW = 1 keV/Nuc) the lowest energy does not fulfill the above
requirement.
Computed with approximation
Computed with full relation
=2
=4
=2
=4
A
A
A
A
10
1.26E+00
2.53E+00
8.64E-01
9.89E-01
100
4.00E-01
8.00E-01
3.81E-01
6.66E-01
1000
1.26E-01
2.53E-01
1.26E-01
2.48E-01
The anisotropy is most pronounced for the low energy ions and becomes less important for high
energies, because the change in energy due to the transformation between the two frames
becomes a small modification for high energies. In the same way it is expected that the influence
of the Compton-Getting Effect on the spectral slope is strongest at low energies. In fact, the
approximation is not good at all at low energies. A always has to be < 1!!
E in keV/Nuc
Phys 954, Solar Wind and Cosmic Rays
E. Möbius
The center curve is the original spectrum in the plasma frame, while the upper and lower curve
represent the two transformed spectra in the upstream and downstream direction as viewed by an
observer at rest. Indeed the slope is changed the most at low energies.
b)
As can be seen from a) the decision on a representation should in principle be made for
the distribution in the plasma rest frame, because the transformation causes significant deviations
from the original slope. Also should we consider the high energies rather than the low energies.
E in keV/Nuc
(in S/C frame)
ju(E) in
jd(E) in
jav (E) in
1/(s cm2 sr keV/Nuc)
1/(s cm2 sr keV/Nuc)
1/(s cm2 sr keV/Nuc)
10
4.57E+05
3.33E+04
2.45E+05
32
2.19E+04
5.20E+03
1.35E+04
100
1.52E+03
6.83E+02
1.10E+03
316
1.26E+02
8.03E+01
1.03E+02
1000
1.14E+01
8.83E+00
1.01E+01
3162
1.07E+00
9.32E-01
1.00E+00
10000
1.04E-01
9.61E-02
1.00E-01
31623
1.02E-02
9.78E-03
1.00E-02
100000
1.01E-03
9.87E-04
1.00E-03
To approximate the original spectrum the average of the upstream and downstream fluxes, as
given in the energy E in the S/C frame, is taken in the last column. From the fluxes above 100
keV/Nuc it can be seen that a power law is the better representation. Taking the points from
1000-100,000 keV/Nuc the power index can be computed directly to =2.
The anisotropies are: (A = 2(Esw/E)1/2 )
A = 0.38
for E = 100 keV/Nuc
and
A = 0.04
for E = 10,000 keV/Nuc
These are good values, as derived from the approximate relation for A, as can be seen in
comparison with those calculating A directly from ju and jd in the data as A = (ju-jd)/(ju+jd)
c)
To test for isotropy we have to evaluate the measured distribution at energies in the
upstream and downstream direction, which correspond to the same energy (E’u,d = 10, 100, ...
keV/Nuc) in the plasma rest frame. I.e. we have to derive the fluxes at the energies Eu and Ed in
the S/C frame, where
Eu = m/2 ((2E’/m)1/2 + VSW)2; Ed = m/2 ((2E’/m)1/2 - VSW)2 and ju,d(Eu,d) = ju,d(E)* (Eu,d/E)-
Then the fluxes have to be extrapolated from the observed fluxes ju,d(E). The transformed
energies Eu,d and fluxes ju,d(Eu,d) and the original fluxes ju,d(E) are compiled in the table.
E' in keV/Nuc
10
32
100
316
1000
3162
10000
31623
100000
ju(E) in
Eu
ju(Eu)
jd(E) in
Ed
4.57E+05
17.3
1.52E+05
3.33E+04
2.19E+04
44.3
1.14E+04
5.20E+03
1.52E+03
121.0
1.04E+03
6.83E+02
1.26E+02
352.6
1.01E+02
8.03E+01
1.14E+01
1064.2
1.01E+01
8.83E+00
1.07E+00
3275.5
9.97E-01
9.32E-01
1.04E-01
10201.0
9.99E-02
9.61E-02
1.02E-02
31979.7
9.97E-03
9.78E-03
1.01E-03 100633.5
9.97E-04
9.87E-04
jd(Ed)
4.7
21.7
81.0
281.4
937.8
3050.5
9801.0
31268.3
99368.5
1.52E+05
1.13E+04
1.04E+03
1.01E+02
1.00E+01
1.00E+00
1.00E-01
1.00E-02
1.00E-03
Phys 954, Solar Wind and Cosmic Rays
E. Möbius
The distribution is obviously isotropic to a very good accuracy. We have achieved such an
excellent agreement, because the original distribution in the SW is indeed a power law
distribution with  = 2, which we derived from the high energy points, as we should. Had we
used the slope found from the observation at low energies, the extrapolation would not have been
good any more at the lowest energies where the slope of the spectrum is heavily affected by the
Compton-Getting Effect.
d)
We
Assume an exponential energy spectrum in the solar wind frame
j( E )  jo e  E / E o
measure the flukes at equal energy E' in the S/C frame, i.e., we in fact determine the
fluxes in upstream and downstream direction for
E'  E 2 2E / m Vsw  mVsw2 / 2
j' E' u d  joe E / Eo e 2 2E / mVSW Eo e
2
A
j' (E' )u  j' (E' )d e

j' (E' )u  j' (E' ) d e

mVSW 2
2Eo
2E
2E
V /E
2
V /E
m SW o  e
m SW o

e
A tanh 2 2E' / m vSW / Eo 
The anisotropy is small for low energies and would be:
E:
E0
A0
if the transition to E = 0 would be a valid approximation However, we measure in the frame of
E', i.e. we need let: E'  0, which is not identical with E!
On the other hand A reaches its maximum
E'  
asymptotically for
A1
i.e. for an exponential spectrum the anisotropy is large at high energies. This becomes obvious,
when we note that an exponential spectrum becomes steeper and steeper at an exponential pace
towards higher energies. Therefore, the Compton-Getting Effect becomes ever stronger.
Phys 954, Solar Wind and Cosmic Rays
3) Essay
E. Möbius
Phys 954, Solar Wind and Cosmic Rays
E. Möbius
2) Diffusion-Convection Equation
a) For u and D independent of r the diffusion convection equation reads:
n
2
 un  Dn
{D n}
t
The main trick is to turn this equation into an ordinary diffusion equation by taking a
transformation into the frame moving with u, i.e. r' = r - ut
Then:
n(r' ,t)
 Dn(r' ,t)  Q(ro,t)
t
where Q(ro,0) = No (r') is an instantaneous injection at time 0, i.e. here a delta function in
space and time, at ro = 0, t = 0. Otherwise we can write in spherical coordinates:
n(r' ,t)
1 
n(r' ,t)
 2
(r' 2 D
)  0
t
r'
r' r'
This can be solved with the convected Gaussian in Kallenrode:
No
n(r' ,t) 
exp(
r' 2
)
4Dt

No
(4Dt )3
(4Dt )3
Inserting into the equation shows that this solution fits.
exp(
(r  ut)2
)
4Dt
One way to arrive at this solution is the use of a Laplace Transformation (use of a Green's
Function Method works as well):

1
ikr'
3/ 2  n(r' ,t)e dr'
(2)
0
The time derivative is not affected by the integral, and the r-derivatives transform into:
 k 2n˜(k,t )
making the differential equation into:
Laplace Transformed n˜ (k,t) 
dn˜ (k,t)/ dt   Dk 2n˜(k,t) with the solution
2
No
Dk 2t
n˜(k,t )  n˜ (k,0) e Dk t 
e
(2)3 / 2
The return transformation leads to:
No  Dk 2t ikr' 3
n(r' ,t) 
e
d k
 e
(2)3 
2
2
where  (Dk t  ikr' )  (Dtk  2(
Thus
After
n(r' ,t) 
No
e
(2)3
(
ir'
r' 2
r' 2
)
)
2 Dt)
4Dt
4Dt
r' 2 
)
2
2
4Dt
e ( Dtk ikr' ) d 3k


Phys 954, Solar Wind and Cosmic Rays
E. Möbius
Dt k  k'
the triple integral results in

2Dt
3
and thus
n(r' ,t) 
No
4Dt
(
3e
r' 2
)
4Dt

No
4Dt

3e
(r ut) 2
4Dt
This differential equation can also be solved by separation of the r’ and t variables after the
transformation. A simple exp ansatz for time will do the trick as in the corresponding problem in
HW set 5. The technique shown here is an alternative one that can be used generally for such an
initial value problem.
Phys 954, Solar Wind and Cosmic Rays
E. Möbius
b) The 3-dimensional diffusion coefficient (sorry, I used here D for the spatial coefficient, while
I used D for the momentum diffusion in class) is:
D =v/3
where  is the mean free path (0.1 AU) and v = Vsw*√E/Esw = 4.4 105*√30 m/s.
Then we get a diffusion coefficient:
D = 0.1*1.5 1011 4.4 105√30 m2/s = 1.2 1016 m2/s
n(r' ,t) 
No
exp(
(1.51011 r[AU] 4.4 10 5  t)2
)
16
4 1.2 10 t
(4 1.21016 t)3
In the following chart we calculate the density as a function of time at 1 AU. Similarly the spatial
distribution at a fixed time can be displayed.